What's the derivation of an orbiting object's energy in Kerr Spacetime

Question

If in Schwarzschild spacetime, the energy per unit mass ε of an object is given by the following equation in timelike geodesics (derived from the squared magnitude of timelike tangent vectors at θ=π/2): $$\varepsilon^2=\left(\frac{dr}{d\tau}\right)^2+c^2-c^2\frac{r_s}{r}+\frac{\ell^2}{r^2}-\frac{r_s\ell^2}{r^3}$$ then what would the corresponding derivation of the following Kerr orbit equation be? $$\frac{1}{2}(\varepsilon^2-1)=\left(\frac{dr}{d\tau}\right)^2+\frac{\ell^2-a^2(\varepsilon^2-1)}{2r^2}-\frac{M(\ell-ae)^2}{r^3}-\frac{M}{r}$$ Where ε is energy per unit mass and cursive l is the object's angular momentum per unit mass.

It's worth noting that that the Schwarzchild equation is written in natural units whereas the Kerr equation is written in planck units so the values of -1 could represent -c^2.

The second equation comes from a 2015 paper from Stockholm University: The Angular Momentum of Kerr Black Holes (p: 24, eqs: 4.45 & 4.46) http://3dhouse.se/ingemar/exjobb/The%20Angular%20Momentum%20of%20Kerr%20Black%20Holes.pdf

Any derivations of ε in Kerr spacetime would be helpful.

Answer 1

Kerr has two Killing vector fields:

$$T^\mu = (-1,0,0,0)$$ $$L^\mu = (0,0,0,1)$$

related to the time translated and rotational symmetry. Consequently,

We have two constants of motion along a geodesic orbits, if

$$u^\mu = (\frac{dt}{d\tau},\frac{dr}{d\tau},\frac{d\theta}{d\tau},\frac{d\phi}{d\tau}),$$

they are the (specific) energy

$$\epsilon = T_\mu u^\mu = (1+\frac{2M r(r^2+a^2)}{(r^2-2Mr+a^2)(r^2+a^2\cos^2\theta )})\frac{dt}{d\tau}+ \frac{2aM r}{(r^2-2Mr+a^2)(r^2+a^2\cos^2\theta )}\frac{d\phi}{d\tau} $$

and the (specific) axial angular momentum

$$ \ell = L_\mu u^\mu=- \frac{2aM r}{(r^2-2Mr+a^2)(r^2+a^2\cos^2\theta )}\frac{dt}{d\tau}- (\frac{1}{\sin^2\theta(r^2+a^2\cos^2\theta )}-\frac{a^2}{(r^2-2Mr+a^2)(r^2+a^2\cos^2\theta )})\frac{d\phi}{d\tau}. $$

We also no that the norm of the 4-velocity $u^\mu$

$$u^\mu g_{\mu\nu} u^\nu =-1$$

is constant along the orbit.

Taking these three equations, and specializing to the equatorial $\theta=\pi/2$ case we can solve for $ \frac{dr}{d\tau}$

to find

$$\left( \frac{dr}{d\tau}\right)^2 = \frac{(\epsilon(r^2+a^2)-a\ell)^2 - (r^2-2Mr+a^2)(r^2+(a\epsilon-\ell)^2)}{r^4} $$